Consider the evaporation of methanol at 25.0 C: CH3OH(l)CH3OH(g). Why methanol s
ID: 1020198 • Letter: C
Question
Consider the evaporation of methanol at 25.0 C: CH3OH(l)CH3OH(g).
Why methanol spontaneously evaporates in open air at 25.0 C: Methanol evaporates at room temperature because there is an equilibrium between the liquid and the gas phases. The vapor pressure is moderate (143 mmHg at 25.0 degrees centigrade), so a moderate amount of methanol can remain in the gas phase, which is consistent with the free energy values.
Find G at 25.0 C under the following nonstandard conditions:
(iii) PCH3OH= 14.0 mmHg .
Express your answer using two significant figures.
I've been able to find it at other pressures but cannot seem to figure out why this pressure says its inccorect if I do it the same way.
Explanation / Answer
Find ?G° at 25 °C. The superscript "°" means that you are considering the conditions T = 298 K, P = 1 atm and you must use the "?" or write "deltaG".
CH?OH(?) ? CH?OH(g)
You also never gave us any information, and we can't answer your question without some information. In this case, I looked up the ?G's of formation for you (your text may have slightly different numbers):
?G°_f(CH?OH(?)) = -166.36 kJ/mol, ?G°_f(CH?OH(g)) = -162.00 kJ/mol
Therefore, ?G°(vap) = -162.00 - (-166.36) kJ/mol = +4.4 kJ/mol
Since ?G° is positive, that means that if the vapor pressure of methanol at 298 K were 1.0 atm, the equilibrium will run the other way, that is, it will condense to form more liquid until it reaches the equilibrium vapor pressure.
b) ?G = ?G° + RTlnQ . . . in this case Q = P/P?, P is the vapor pressure and P? = 1 atm
i. ?G = ?G° + RTlnQ = +4.4 kJ/mol + (8.314 J/mol•K)(298 K)ln(150 mm/760 mm)
?G = +4.4 - 4.0 kJ = +0.4 kJ/mol
ii. ?G = ?G° + RTlnQ = +4.4 kJ/mol + (8.314 J/mol•K)(298 K)ln(100 mm/760 mm)
?G = +4.4 - 5.0 kJ = -0.6 kJ/mol
iii. ?G = ?G° + RTlnQ = +4.4 kJ/mol + (8.314 J/mol•K)(298 K)ln(10 mm/760 mm)
?G = +4.4 - 10.7 kJ = -6.3 kJ/mol
The equilibrium vapor pressure (when ?G = 0) is clearly less than 150 torr and greater than 100 torr at 298 K. In case (i), more vaper will condense, in case (ii), more will evaporate - and even more in case (iii)..